Electric field from doorknob

Because the doorknob is a conductor, the entire charge is distributed on the outside surface of the metal. In addition, because the doorknob is assumed to be perfectly spherical, the charge on the surface is uniformly distributed, so we can treat the doorknob as if all the charge were located at the center of the doorknob. The validity of this simplification will be proved in a later physics course. Now sketch the doorknob, and define your coordinate system. Use $+x$ to indicate the outward direction perpendicular to the door, with $x=0$ at the center of the doorknob (as shown in the figure below).

If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. We want to know the electric field 1.0 cm from the surface of the doorknob, which is a distance $r=2.5\text{cm}+1.0\text{cm}=3.5\text{cm}$ from the center of the doorknob. We can use the equation $E=\frac{k\left|Q\right|}{r^2}$ to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge, which is negative in this case.

Inserting the charge $Q=\mathrm{-1.5}\text{nC}=\mathrm{-1.5}\times {10}^{\mathrm{-9}}\text{C}$ and the distance $r=3.5\text{cm}=0.035\text{m}$ into the equation $E=\frac{k\left|Q\right|}{r^2}$ gives

Because the charge is negative, the electric-field lines point toward the center of the doorknob. Thus, the electric field at $x=3.5\text{cm}$ is $(-1.1\times {10}^4\text{N/C})\widehat{x}$ .